Questions about a small memory allocation program (41 lines)

Question

Questions

1. Does the "memory storage array" allocbuf[10] save everything back to back. For example allocbuf[10] is full with {F, i, r, s, t, \0, S, e, c, \0] and I'll state how allocbuf became full. I imagine all of the information I save into allocbuf saves back to back. So, as an example, after using alloc(6), and assigning the returned character pointer to a variable char *first, the character pointer "first" points to allocbuf[0]. Now I assign, first = "First"; If I want to print "First", do I have to use a for loop, for (i = 0; i < 6; i++) { print allocbuf[i]; } to print out "First"?

2. If I assign first = alloc(1) and then assign first = "First", it will work, and overwrite the contents of allocbuf, correct?

3. Why does the line printf("allocbuf: %s\n", allocbuf); not print out the contents of allocbuf in my code?

I'm confident I'll be interested in more features of this program and I am excited to have your help.

I am excited to read any comments about this program, the functions, and memory allocation, although they may not answer on of the specific questions I have. So please share you knowledge and experience with me. Thanks :)

P.S. I haven't encountered malloc() in my book, K&R C Programming 2nd ed, yet, so, please, no comments saying use malloc().

Code

char *alloc(int); //return pointer to free storage in allocbuf[10]
void afree(int *); //free storage in allocbuf[10]
void strcpy2(char *, char *); //copy to, from
static char allocbuf[10];
static char* allocp = allocbuf;

main()
{
    char array4[5] = "4444";
    char* cp = "overwritten";
    char* copy = array4;
    char* occupyalloc;

    printf("cp: %s\n", cp); //"overwritten"
    printf("copy: %s\n", copy); //"4444"

    cp = "2"; //overwrite *cp = "overwritten" with *cp = "2"
    printf("cp: %s\n", cp); //"2"

    occupyalloc = alloc(4); //returns allocp 0, intended storage space is allocbuf[0] through allocbuf[3]

    cp = alloc(3); //returns allocp 4, intended use is allocbuf[4] through allocbuf[6]

    strcpy2(cp, copy); //copies "4444" into *cp, specifically allocbuf[4] through allocbuf[7]

    printf("cp: %s\n", cp); //"4444" , stored in allocbuf[4] through allocbuf[7] improperly
    printf("allocbuf: %s\n", allocbuf); //prints allocbuf -- not working
}

char *alloc(int n)
{
    if (allocbuf + ALLOCSIZE - allocp >= n) 
    {
        allocp += n;
        return allocp - n;
    }
    else
        return 0;
}

void afree(int *initial_storage_element_location)
{
    if (initial_storage_element_location >= allocbuf && initial_storage_element_location < allocbuf + ALLOCSIZE) 
        allocp = initial_storage_element_location;
}

void strcpy2(char *s, char *t)
{
    while (*s++ = *t++)
            ;
}

Answer 1

You should understand that if you have a line as cp = "2" then

• The compiler creates a string literal as "2".
• This string literal is generally stored in the read only location of the memory.
• The address of the string literal is stored in the pointer cp

Now, if you are using dynamic memory allocation as

 char *first = alloc(6);
 first = "First";

What happens in the first statement is that a memory is allocated from the static buffer allocbuf and the address of the allocated memory is stored in first. With the second statement, you are overwriting the pointer first with a location in the read only memory. So you loose the previously allocated pointer and you have memory leakage

If you are using alloc you should copy the data First into the memory with strcpy

strcpy(first, "First")

If I assign first = alloc(1) and then assign first = "First", it will work, and overwrite the contents of allocbuf, correct?

No, you not overwriting the contents of allocbuf but you are using a new pointer to read only memory.

Why does the line printf("allocbuf: %s\n", allocbuf); not print out the contents of allocbuf in my code?

I will answer this in the context of the code posted.

occupyalloc = alloc(4); //returns allocp 0, intended storage space is allocbuf[0] through allocbuf[3]
cp = alloc(3); //returns allocp 4, intended use is allocbuf[4] through allocbuf[6]
strcpy2(cp, copy); //copies "4444" into *cp, specifically allocbuf[4] through allocbuf[7] 

printf("cp: %s\n", cp); //"4444" , stored in allocbuf[4] through allocbuf[7] improperly printf("allocbuf: %s\n", allocbuf); //prints allocbuf -- not working

In this case, allocbuf[0] will be \0, as you are not modifying occupyalloc variable.

Note also that "4444" requires 5 bytes and you are only assigning 3 bytes.